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Double Pointers - Question 1

167. Two Sum II - Input Array Is Sorted

Given a 1-indexed array of integers numbers that is already sorted in non-decreasing order, find two numbers such that they add up to a specific target number. Let these two numbers be numbers[index1] and numbers[index2] where 1 <= index1 < index2 <= numbers.length.

Return the indices of the two numbers, index1 and index2, added by one as an integer array [index1, index2] of length 2.

The tests are generated such that there is exactly one solution. You may not use the same element twice.

Your solution must use only constant extra space.


Constraints:

2 <= numbers.length <= 3 * 104
-1000 <= numbers[i] <= 1000
numbers is sorted in non-decreasing order.
-1000 <= target <= 1000
The tests are generated such that there is exactly one solution.


Analysis:

The array is sorted, so we may want to consider binary search, but actually we can use a two-pointer method to quickly solve this problem, which is faster than the binary search method.

The first pointer (p1) points to the start, and the other (p2) points to the end.
At each step, we only have three possible cases, which are listed below

while (p1 < p2)
    if numbers[p1] + numbers[p2] == sum
        return {p1, p2};
    else if numbers[p1] + numbers[p2] < sum
        ++p1;
    else
        --p2;

Since we just scan the array (at most) once, the time complexity is O(N), where N is the array size. The space complexity is O(1).

If we use binary search, we need to fix the first element first, then do a binary search. So the overall time complexity is O(Nlog(N)).


See the code below:

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        int x = 1, y = (int)numbers.size();
        while(x < y) {
            int sum = numbers[x-1] + numbers[y-1];
            if (sum == target) {
                return {x, y};
            } else if (sum < target) {
                ++x;
            } else {
                --y;
            }
        }
        return {-1, -1};
    }
};



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